Prove $x_n=\frac{n}{2^n}$ converges to $0$ Prove $x_n=\frac{n}{2^n}$ converges to $0$.
Since $2^n>\frac{n^2}{2}$ for all $n \in \mathbb{N}$ we have
$$-\frac{2}{n}<\frac{n}{2^n}<\frac{2}{n}.$$
Since $\frac{1}{n} \rightarrow0$ as $n \rightarrow \infty$, by Squeeze theoreom $\frac{n}{2^n}$ converges to $0$.
Is there anything wrong this proof?
 A: Your proof is perfectly fine just that you don't need $\frac{-2}{n}$. You can just replace that by $0$.
Here's another alternative method by using the ratio test:
$\frac{x_{n+1}}{x_n}=\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}=\frac{1}{2}\left(\frac{n+1}{n}\right)$.
Now take the limit on both sides and thus, using the theorem of ratio test for sequences, if the limit $L$ exists and $0\leq L<1$ then $\text{lim }x_n=0$.
A: $$\frac n{2^n}=\frac12\frac2{1\cdot2}\frac3{2\cdot2}\frac4{3\cdot2}\frac5{4\cdot2}\cdots\frac{n-1}{(n-2)2}\frac n{(n-1)2}
\\<\frac12\frac2{1\cdot2}\frac3{2\cdot2}\frac3{2\cdot2}\frac3{2\cdot2}\cdots\frac3{2\cdot2}\frac3{2\cdot2}\to0$$

Possibly more obvious:
$$n=1\frac21\frac32\frac43\frac54\cdots\frac{n-1}{n-2}\frac n{n-1}<1\frac21\frac32\frac32\frac32\cdots\frac32\frac32=2\left(\frac32\right)^{n-2}.$$
And of course
$$\frac n{2^n}<\frac12\left(\frac34\right)^{n-2}.$$
A: Your proof seems legit. However,  your problem can be easy solved using L'Hospital Rule.
$\lim\limits_{n\to\infty}\frac{n}{2^n}=\lim\limits_{n\to\infty}\frac{1}{2^n\log2}=0$
